I'm struggling with probability questions, even those that should be easy! :( We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).

The question is:

Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?

My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA

My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?

Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?

reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.

d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?

Please help! Thanks!

Oct 8th 2009, 03:58 AM

mr fantastic

Quote:

Originally Posted by pianopiano

I'm struggling with probability questions, even those that should be easy! :( We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).

The question is:

Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat? Mr F says: The correct answer is 2/6 (BCA and CAB). Read Derangement - Wikipedia, the free encyclopedia
b) That exactly 1 man will? Mr F says: 3/6. So you are correct.
c) That exactly 2 men will? Mr F says: 0. If 2 men out of the 3 get a correct hat, the remaining man must also get a correct hat.
d) That all 3 will? Mr F says: 1/6. So you are correct.

My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA

My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?

Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?

reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.

d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?

Please help! Thanks!

Note that all the probabilities have to add up to 1. 2/3 + 1/2 + 1/6 > 1 so you know that there's got to be mistakes (your answer to (a) was wrong).